This document contains lecture notes from Physics 1301 on friction. It discusses static and kinetic friction, the minimum slope angle needed for skiing, and provides examples of problems involving boxes on surfaces with friction. It also covers stopping distance calculations for cars using static friction between tires and the road.
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This document contains lecture notes from Physics 1301 on friction. It discusses static and kinetic friction, the minimum slope angle needed for skiing, and provides examples of problems involving boxes on surfaces with friction. It also covers stopping distance calculations for cars using static friction between tires and the road.
This document contains lecture notes from Physics 1301 on friction. It discusses static and kinetic friction, the minimum slope angle needed for skiing, and provides examples of problems involving boxes on surfaces with friction. It also covers stopping distance calculations for cars using static friction between tires and the road.
Copyright:
Attribution Non-Commercial (BY-NC)
Available Formats
Download as PPT, PDF, TXT or read online from Scribd
This document contains lecture notes from Physics 1301 on friction. It discusses static and kinetic friction, the minimum slope angle needed for skiing, and provides examples of problems involving boxes on surfaces with friction. It also covers stopping distance calculations for cars using static friction between tires and the road.
Copyright:
Attribution Non-Commercial (BY-NC)
Available Formats
Download as PPT, PDF, TXT or read online from Scribd
Download as ppt, pdf, or txt
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Physics 1301: Lecture 16, Pg 1
Lecture 16: More on Friction
u a T m u =30 0 N mg f S a
v o mg Physics 1301: Lecture 16, Pg 2 Friction Review Surface friction is caused by the microscopic interactions between the two surfaces: A surface exerts a force on a body placed on it We can decompose this force into the normal force vector N and the frictional force vector f F
perpendicular to N and parallel to the surface, in the direction opposing relative motion of the two surfaces. Kinetic (sliding): The magnitude of the frictional force is proportional to the magnitude of the normal force N. f K = u K N Static: The frictional force balances the net applied forces such that the object doesnt move. The maximum possible static frictional force is proportional to N. f S s u S N
Physics 1301: Lecture 16, Pg 3 Skiing What is the minimum slope angle needed to ski? Use u k = 0.04 for waxed wood on dry snow. u a Physics 1301: Lecture 16, Pg 4 u mg
Skiing x k x k k y ma mg mg ma N mg f mg x ma mg N y = = = = = u u u u u u u cos sin sin sin : 0 cos : What is the minimum angle needed to ski? Draw Free Body Diagram Choose axes, parallel and perpendicular to the slope
> > > > 2 04 . 0 tan tan . . cos sin u u u u u u u So e i mg mg Need k k Skier will slide if acceleration > 0 Physics 1301: Lecture 16, Pg 5 Hiking What about rubber on concrete? Use u k = 0.8 ! ! 39 8 . 0 tan > > u u So Need slipping start to 45 So 0 . 1 tan > > u u Need Static case u s = 1.0 mg Physics 1301: Lecture 16, Pg 6 Problem: Box on Truck A box with mass m sits in the back of a truck. The coefficient of static friction between the box and the truck is u S . What is the maximum acceleration a that the truck can have without the box slipping? m u S a Physics 1301: Lecture 16, Pg 7 Problem: Box on Truck Draw a Free Body Diagram for the box:
Consider case where f F is max... (i.e. if the acceleration were any larger, the box would slip). Then the friction force is the maximum static friction f F =u s N N f F = u S N mg x y Physics 1301: Lecture 16, Pg 8 Problem: Box on Truck Use F NET = ma for both x and y components N f F = u S N mg a MAX x y x u S N = ma MAX
y N = mg a MAX = u S g Physics 1301: Lecture 16, Pg 9 Problem Box and Rope A box of mass m =10.21 kg is at rest on a floor. The coefficient of static friction between the floor and the box is u s = 0.4. A rope is attached to the box and pulled at an angle of u = 30 o above horizontal with tension T = 40 N. Does the box move?
T m
static friction (u s = 0.4 ) u UIUC Physics 1301: Lecture 16, Pg 10 Solution Box and Rope Pick axes & draw FBD of box: Apply F NET = ma y: N + T sin u - mg = ma Y = 0
N = mg - T sin u = 10.21 kg 9.81 m/s 2 - 40N 0.5 = 80 Newtons x: T cos u - f s = ma X y x T m u =30 0 N mg f S Physics 1301: Lecture 16, Pg 11 Solution Box and Rope x: T cos u - f s = ma X y: N = 80 N
The box will move if T cos u - f S > 0 T cos u = 40cos30 = 34.6 N f s = u s N = (.4)(80N) = 32 N So T cos u > f s The box does move T m u=30 0
N mg f S Physics 1301: Lecture 16, Pg 12 ICQ: Box and Rope A box of mass m is pulled in the direction shown in the figure across a rough surface at a constant velocity. The magnitude of the frictional force is a) u k mg b) u k Tcosu c) u k (T-mg) d)u k Tsinu e)u k (mg-Tsinu) T u m Physics 1301: Lecture 16, Pg 13 ICQ: Box and Rope A box of mass m is pulled in the direction shown in the figure across a rough surface at a constant velocity. The magnitude of the frictional force is a) u k mg b) u k Tcosu c) u k (T-mg) d)u k Tsinu e)u k (mg-Tsinu) The frictional force is F = u k N The normal force is given by N + Tsinu = mg So F=u k (mg-Tsinu) T u F mg N Physics 1301: Lecture 16, Pg 14 Stopping Distance
A car is moving with initial speed v o . The brakes are applied. The coefficient of static friction between the wheels and the road is u S . Find the stopping distance D. a
v o v = 0 D Physics 1301: Lecture 16, Pg 15 Stopping distance 0 = = = = v g a mg N ma s s s u u u f F
Ax g v x s u 2 2 0 = A Stopping distance x a v v A = 2 2 0 2 Use Force stopping car is static friction between tire and road This is why the police measure the skid marks on the road They can tell if you were exceeding the speed limit Physics 1301: Lecture 16, Pg 16 Stopping distance Examples For kinetic friction first, if the car is sliding
Now for static friction if wheels are rolling m g v s m v For k 25 8 . 9 8 . 0 2 400 2 1 x 8 . 0 , / 20 2 =
= = A = = u u f F
Ax m g v s m v For s 20 8 . 9 0 . 1 2 400 2 1 x 0 . 1 , / 20 2 =
= = A = = u u These stopping distances will be much longer on ice (u S =0.1) Physics 1301: Lecture 16, Pg 17 Motive force of your car What is the force that provides the acceleration of your car? a) The force of your foot on the accelerator b) The force of burning gas on the piston in the engine c) The force of the engine turning the wheels d) The force of friction between the tires and the road. Physics 1301: Lecture 16, Pg 18 Motive force of your car What is the force that provides the acceleration of your car? a) The force of your foot on the accelerator b) The force of burning gas on the piston in the engine c) The force of the engine turning the wheels d) The force of friction between the tires and the road. Physics 1301: Lecture 16, Pg 19 Tires are your cars only contact with the road. All the forces that cause your car to accelerate or brake are transmitted through the tires Your engine does not accelerate the car, it turns the wheels that accelerate the car. The point of contact of your tire with the road is at rest, provided you are not skidding
It is the force of static friction between the tires and the road that provides the force to accelerate or brake your car Anti-lock brakes work by making sure the wheels roll without slipping. This maximizes the frictional force slowing the car since u S > u K .
Tires v v v f Physics 1301: Lecture 16, Pg 20 Problem Two boxes A box of mass m 1 = 1.5 kg is being pulled by a horizontal string having tension T = 90 N. It slides with friction (u k = 0.51) on top of a second box having mass m 2 = 3 kg, which in turn slides on a frictionless floor. What is the acceleration of the second box ?
m 2 T m 1
slides with friction (u k =0.51 ) slides without friction a = ? Physics 1301: Lecture 16, Pg 21 Solution Two Boxes First draw FBD of the top box: m 1 N 1,2 m 1 g T f = u K N 1,2 = u K m 1 g Physics 1301: Lecture 16, Pg 22 Solution Two Boxes Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. There are two forces on box 2 due to box 1 Friction force f 2,1 Contact force N 2,1 f 2,1 f 1,2 = u K m 1 g N 1,2 N 2,1 m 1 m 2
Physics 1301: Lecture 16, Pg 23 Solution Two boxes Now consider the FBD of box 2:
N 2,f = m 1 g+m 2 g =(m 1 +m 2 )g m 2 g N 2,f m 1 g=N 2,1 m 2 f 2,1 = u k m 1 g Physics 1301: Lecture 16, Pg 24 Solution Two Boxes Finally, solve F = ma in the horizontal direction: m 2 f 2,1 = u K m 1 g u K m 1 g = m 2 a g m m a k 2 1 u = 2 s m 81 9 51 0 kg 3 kg 5 1 . . . = a = 2.5 m/s 2
Physics 1301: Lecture 16, Pg 25 Homework
Tomorrow we will learn about circular motion, read Fishbane chapter 5-4 Do problems, chapter 5 #8,9,33,34,